The oxidation state of nickel is 0. In the ground state, nickel has the valence shell electronic configuration 3d8 4s2. However, due to the presence of strong field CO ligands, the two 4s electrons are forced to pair up in the 3d orbitals. As a result, all ten electrons occupy the 3d orbitals, leading to the formation of sp3 hybridisation.
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Questions
Using valence bond theory, predict the hybridization and magnetic character of the complex:
[Ni(CO)4] (Atomic number: Ni = 28)
Using valence bond theory, explain the hybridisation and magnetic character of the following:
[Ni(CO)4] (At. no.: Ni = 28)
Diagram
Short Answer
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Solution
Orbitals of Ni:
| ↑ ↓ | ↑ ↓ | ↑ ↓ | ↑ | ↑ | ↑ ↓ | |||||
| 3d | 4s | 4p | ||||||||
sp3 hybridised orbitals of Ni:
| ↑ ↓ | ↑ ↓ | ↑ ↓ | ↑ ↓ | ↑ ↓ | |||||
| 3d | sp3 hybrid | ||||||||
[Ni(CO)4] (low spin complex):
| ↑ ↓ | ↑ ↓ | ↑ ↓ | ↑ ↓ | ↑ ↓ | XX | XX | XX | XX | |
| 3d | sp3 hybrid | ||||||||
Therefore, the complex [Ni(CO)4] involves sp3 hybridisation and the complex is diamagnetic, as evident from the absence of unpaired electrons.
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